Question

Решите 2 задание а)x^2-5x-2 - 24/x^2-5x=0
б)x^2+3x-3/x^2-4x-7 + x^2-4x-7/x^2+3x-3 = -2
и напишите , максимально понятно.

Answer 1

а)$x^{2} -5x-2-\frac{24}{x^{2} -5x}=0$

ОДЗ: $x^{2} -5x\neq 0; x( x-5)\neq 0$ ⇒ $x\neq 0, x\neq 5$

Замена: $t = x^2-5x$

$t-2-\frac{24}{t} =0|*t\\t^2-2t-24=0$

$$\left[\begin{gathered}t=6 \\ t=-4\end{gathered} \right.$$  $$\left[\begin{gathered}x^2-5x=6 \\ x^2-5x=-4\end{gathered} \right.$$$$\left[\begin{gathered}x^2-5x-6=0 \\ x^2-5x+4=0\end{gathered} \right.$$

$$\left[\begin{gathered} $\left[\begin{gathered} x=6\\x=-1\end{gathered} \right.$\\$\left[\begin{gathered} x=4\\x=1\end{gathered} \right.$\end{gathered} \right.$$

Тогда общий ответ с учётом ОДЗ: $x_1 = -1, x_2 = 1,x_3 = 4, x_4 = 6$.

б)$\frac{x^2+3x-3}{x^2-4x-7} +\frac{x^2-4x-7}{x^2+3x-3}=-2$

Замена: $p=x^2+3x-3; q=x^2-4x-7$

$\frac{p}{q}+ \frac{q}{p}+2=0\\\frac{p^2+q^2}{pq}+2=0\\\frac{p^2+2pq+q^2}{pq}=0\\\frac{(p+q)^2}{pq}=0$

$p+q=x^2+3x-3+x^2-4x-7=2x^2-x-10$

$\frac{(2(x-2,5)(x+2))^2}{pq} \\$\left[\begin{gathered}x=-2 \\ x=2,5\end{gathered} \right.$,pq\neq 0$  

$$\left[\begin{gathered}p\neq 0 \\q\neq 0\end{gathered} \right.$$  $$\left[\begin{gathered}x^2+3x-3\neq 0 \\x^2-4x-7\neq 0\end{gathered} \right.$\\$$$\left[\begin{gathered}x\neq \frac{-3\pm\sqrt{21} }{2} \\x\neq 2\pm\sqrt{11} \end{gathered} \right.$\\$

ответ: $x_1 = -2, x_2=2,5$